What is the integral from 0 to pi/4 of 2tan^5(x) ?

The integral from 0 to pi/4 of 2tan^5(x) is -1/2+ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/4 of 2tan^5(x)

\int_{0}^{\frac{π}{4}} 2 tan^{5} (x) d x

- \frac{1}{2} + ln (2)

Decimal Notation

0.19314 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 2 tan^{5} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{\frac{π}{4}} tan^{5} (x) d x

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{4}} (- 1 + sec^{2} (x))^{2} tan (x) d x

Apply u-substitution

: \int_{1}^{2} \frac{( - 1 + u ^{2} ) ^{2}}{u} d u

= 2 \cdot \int_{1}^{2} \frac{( - 1 + u ^{2} ) ^{2}}{u} d u

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u + u^{3}

= 2 \cdot \int_{1}^{2} \frac{1}{u} - 2 u + u^{3} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{1}^{2} \frac{1}{u} d u - \int_{1}^{2} 2 u d u + \int_{1}^{2} u^{3} d u)

\int_{1}^{2} \frac{1}{u} d u = \frac{1}{2} ln (2)

\int_{1}^{2} 2 u d u = 1

\int_{1}^{2} u^{3} d u = \frac{3}{4}

= 2 (\frac{1}{2} ln (2) - 1 + \frac{3}{4})

Simplify

2 (\frac{1}{2} ln (2) - 1 + \frac{3}{4}) : - \frac{1}{2} + ln (2)

= - \frac{1}{2} + ln (2)

x^{2} - 2 x + 42

\frac{1}{7} x, 7 x^{2}

x \to \infty lim ((x^{2} - 1) e^{- x^{2}})

\int - x cos (x + \frac{π}{3}) d x

What is the integral from 0 to pi/4 of 2tan^5(x) ?
The integral from 0 to pi/4 of 2tan^5(x) is -1/2+ln(2)